MATH U113 · Mid-semester exam (07 Oct 2026) · past papers
Three mid-sem papers, every question from zero
The last three mid-sem papers of this course (as MATH F113): 09 Mar 2026 (closed book — the same rules as yours), 03 Mar 2025 and 13 Mar 2024 (open book). Each question: which tool, the steps with every number shown, a figure where the picture carries the argument, the traps, an in-room check, and the answer to four decimal places. Practise the same families with fresh numbers on the mid-sem drill.
All 19 questions come from Modules 1–3 — handout lectures L1–20: probability and Bayes, the discrete models, the continuous models. Not one touches joint distributions (Module 4). The official mid-sem scope has not been announced; it will come from the instructor-in-charge (class or LMS notice), and this page will say so when it does.
Most of it is already on your pages. Bayes trees, "exactly k of n", hypergeometric, Poisson windows, the normal table, continuity correction, the memoryless exponential — all in the Module 1–3 notes. Three things were genuinely missing, and are now written: MGFs of continuous distributions and MGF limits (2025 Q1, Q7 → new Module 3 part 2), and the pmf recurrence (2025 Q6 → Module 2 notes). Those are new to everyone in the hall — no board exam teaches them.
The pattern that costs marks is not difficulty, it is stacking. The biggest questions (2026 Q4, 14 marks; Q6(b), 10 marks) chain three small ideas you already know: a distribution gives a p, the p feeds a binomial or hypergeometric, that feeds Bayes. Each link is a Module 1–3 one-liner. The skill is writing the links down in order.
The three papers
| Paper | Book | Time · marks | Questions | Notes |
|---|---|---|---|---|
| 09 Mar 2026 | Closed | 90 min · 60 | 6 (4, 4, 8, 14, 3+5, 5+10+7) | "Round final answers to four decimal places." Four Φ values printed on the paper — only Φ(1.55) is used by any question. |
| 03 Mar 2025 | Open | 90 min · 60 | 7 (9, 4, 6, 4, 5+8, 12, 12) | Heavier on derivations: an MGF to recognise, a pmf recurrence, an MGF limit. |
| 13 Mar 2024 | Open | 90 min · 60 | 6 (10, 10, 12, 8, 8, 9+3) | Two WhatsApp photos. Someone's handwritten answers are on them — all right except one intermediate line (Q3(b)). Page 2 is the friendliest stretch of all three papers: direct Poisson, normal and pdf calculations with no chaining. |
All three share the same instruction block: define the events and random variables clearly, specify all distributions properly, show all steps. That is where partial credit lives. A line like "Let N = number of failed servers; given Enigma-1, N ~ Bin(10, 0.1175)" earns marks even if the arithmetic after it slips.
Topic × paper
| Topic | 2026 | 2025 | 2024 | Where it lives |
|---|---|---|---|---|
| Conditional probability, total probability, Bayes | Q4, Q6ab | Q5 | Q1, Q2 | M1 §2.4 |
| Binomial — "exactly / at least k of n" | Q4, Q5b, Q6c | Q5b | Q3 | M2 models |
| Hypergeometric (without replacement) | Q6b | Q4 | M2 models | |
| Poisson process | Q5 | Q4 | M2 models | |
| Negative binomial / geometric / "first to i wins" | Q6, Q7 | Q3 | M2 models, recurrence | |
| pdf / cdf / E of a continuous rv; minimising an expectation | Q1, Q2 | Q3b, Q6a | M3 §4.1–4.2 | |
| Normal; normal approximation to the binomial | Q3 | Q1, Q2 | Q5 | M3 §4.3 |
| Exponential / gamma, memoryless | Q4 | Q3 | Q4d, Q6b | M3 §4.4 |
| MGF: recognise a distribution; limits | Q1, Q7 | M3 part 2 |
Not in any paper we have: Chebyshev's inequality and transformation methods (handout L19–20). They are lectured before the mid-sem, and your paper is set by a new instructor team — so they are covered on the Module 3 part 2 pages, not skipped.
Closed book: what must be in your head
Your paper is closed book, like 2026. Expect the Φ values to be printed (they were in 2026); everything else below has to come from memory. Each line is one you will use.
- Bayes: P(Ai | B) = P(B | Ai)P(Ai) / Σj P(B | Aj)P(Aj) — the denominator is total probability.
- Binomial C(n, x)pxqn−x, mean np, variance npq. "At least 2" = 1 − qn − npqn−1.
- Hypergeometric C(M, x) C(N−M, n−x) / C(N, n). Poisson e−μμx/x! with μ = αt (rate × window).
- Negative binomial, both conventions: failures before the r-th success (Devore) C(x+r−1, r−1)prqx, mean rq/p; trials until it C(x−1, r−1)prqx−r, mean r/p.
- Continuous: f = F′, E(X) = ∫ x f(x) dx, V(X) = E(X²) − μ². Uniform(A, B): mean (A+B)/2, variance (B−A)²/12.
- Normal: Z = (X − μ)/σ, Φ(−z) = 1 − Φ(z); MGF eμt + σ²t²/2. Binomial → normal when np ≥ 10 and nq ≥ 10, with the ±0.5 continuity correction.
- Exponential: F(x) = 1 − e−λx, mean 1/λ — "mean 12 000 h" means λ = 1/12 000. Memoryless: P(X > s+t | X > s) = P(X > t). Gamma(α, β): mean αβ, variance αβ², MGF (1 − βt)−α; α = 1 is the exponential; χ²(ν) = gamma(ν/2, 2).
- MGFs: binomial (q + pet)n, Poisson eμ(et−1), and MaX(t) = MX(at). Chebyshev: P(|X − μ| ≥ kσ) ≤ 1/k².
The 90-minute plan (built on the 2026 paper)
60 marks in 90 minutes is 1.5 minutes per mark. Spend about 1.3 and keep the rest for checking. Short, certain questions first; the long chain last, when nothing else is waiting.
| Order | Question | Marks | Minutes | Why here |
|---|---|---|---|---|
| 0 | Read the whole paper; write the family name next to each question | — | 3 | Naming the tool is the real skill |
| 1 | Q6(a) Bayes, two machines | 5 | 5 | One tree, two multiplications |
| 2 | Q5 Poisson regions | 8 | 9 | Two formulas you know |
| 3 | Q1, Q2 continuous E | 8 | 11 | Pure integration, no reading traps |
| 4 | Q3 normal approximation | 8 | 10 | The printed Φ(1.55) confirms your z |
| 5 | Q6(c), then Q6(b) | 17 | 20 | (c) reuses 0.07 from (a); (b) is (a)'s tree with new leaves |
| 6 | Q4 the pipeline | 14 | 20 | Longest chain; worth it only with a clear head |
| 7 | Checks: every probability in [0, 1], posteriors sum to 1, 4 dp | — | 12 | The in-room checks listed under each question |
Paper 1 · 09 Mar 2026 · closed book · 60 marks
“Answer all the questions. Symbols have their usual interpretation as per textbook. Marks to each question are mentioned at its end. Answer all parts of the same question together, showing all steps in detail. Specify all distributions properly. Define the event(s) and random variable(s) clearly. Simplify the answers fully. Round off your final answers upto four decimal places.”
The printed table on this paper: Φ(1.55) = 0.9394, Φ(1) = 0.8413, Φ(0.5) = 0.6915, Φ(1.2) = 0.8849. Final answers to four decimal places.
Q1 · E[X] from a cdf · 4 marks §4.2
Where this lives M3 notes §4.1–4.2 · f = F′, E by integration
Which toolYou are handed the cdf; expectation needs the pdf. Differentiate, then integrate x f(x).
Step 1pdf from cdf
Expand first: F(x) = 2x − x², so on [0, 1] f(x) = F′(x) = 2 − 2x, and 0 elsewhere. Check it is a pdf: f ≥ 0 on [0, 1] and ∫01(2 − 2x) dx = [2x − x²]01 = 1 ✓.
Step 2Integrate x·f(x)
1) Integrating x F(x) instead of x f(x) — gives 5/12, wrong. The cdf is a running total, not a density. 2) Forgetting to expand before differentiating and then slipping on the product rule. x(2 − x) = 2x − x² first; then differentiate term by term.
In-room check. For a variable living on [0, ∞), E[X] = ∫ (1 − F(x)) dx. Here 1 − F = 1 − 2x + x² = (1 − x)², and ∫01(1 − x)² dx = 1/3 ✓. Also, the pdf leans towards 0, so the mean must be below ½ ✓.
Q2 · Guess a uniform number: minimise E|X − g| · 4 marks §4.2
Where this lives M3 notes §4.2 · E[h(X)] = ∫ h(x) f(x) dx
Which toolExpected value of a function of X: E[h(X)] = ∫ h(x) f(x) dx. The answer is a function of g; then ordinary calculus minimises it.
Step 1Split the absolute value where it changes sign
f(x) = 1 on [0, 1]. For x < g, |x − g| = g − x; for x > g, it is x − g:
Each integral is the area of a right triangle with two equal legs (left figure): legs g and g, then 1 − g and 1 − g.
Step 2Minimise over g
Expand: h(g) = g²/2 + (1 − 2g + g²)/2 = g² − g + ½. Then h′(g) = 2g − 1 = 0 at g = ½, and h″ = 2 > 0, so it is a minimum. h(½) = ¼ − ½ + ½ = ¼. Endpoints give h(0) = h(1) = ½, larger ✓.
1) Writing E|X − g| = |E[X] − g| = |½ − g|. Expectation does not pass through an absolute value; that shortcut says the minimum is 0, which is impossible (you are almost never exactly right). 2) Answering only "g = ½" — the question asks for the minimum expected score as well.
In-room check. Guessing ½, the error is uniform on [0, ½] (by symmetry), whose mean is ¼ ✓.
Q3 · Fair or biased coin: the chance of a false conclusion · 8 marks §4.3
Where this lives M3 notes §4.3 · normal approximation to the binomial, continuity correction
Which toolA binomial count with n = 1000 and "approximate" in the question → normal approximation with continuity correction.
Step 1Define, and translate "false conclusion"
Let X = number of heads in 1000 tosses. The coin is fair, so X ~ Bin(1000, 0.5). A fair coin leads to a false conclusion exactly when we call it biased: the event X ≥ 525. (The 55% coin plays no part in this question.)
Step 2Check the approximation and get μ, σ
np = nq = 500 ≥ 10 ✓. μ = np = 500, σ = √(npq) = √250 = 15.8114.
Step 3Continuity correction, standardise, read the table
The bar for 525 covers [524.5, 525.5], so "525 or more" starts at 524.5:
This is why Φ(1.55) is printed on the paper: 1.5495 rounds to 1.55. Without the correction you would get 25/15.81 = 1.58, a value the paper doesn't give you — that mismatch is itself the hint.
1) Using the biased coin's p = 0.55. "If the coin is actually fair" fixes p = 0.5; the 0.55 would only matter for the other error (a biased coin called fair). 2) Dividing by the variance 250 instead of σ = 15.81. 3) Correcting in the wrong direction: "≥ 525" goes down to 524.5, not up to 525.5.
In-room check. 525 is about 1.6σ above the mean; a one-sided tail beyond 1.6σ is about 5–6% ✓. (The exact binomial value is 0.0606 as well.)
Q4 · Server racks: exponential → binomial → Bayes · 14 marks §4.4, 3.4, 2.4
Where this lives M3 notes §4.4 · exponential cdf · M2 · binomial · M1 §2.4 · Bayes · the same "likelihood built by a model" pattern as Tutorial 2 Q5
Which toolA three-link chain. The exponential gives each server's failure probability p; ten independent servers make the count of failures Bin(10, p); "at least 2 failed" is the evidence for Bayes between the two makers.
Step 1Name everything
E1, E2 = "rack from Enigma-1 / Enigma-2", with P(E1) = 0.8, P(E2) = 0.2. T = a server's failure time. N = number of the 10 servers failed by 1 500 h. A = "N ≥ 2". Wanted: P(E2 | A).
Step 2Link 1 — one server's failure probability
Exponential with mean θ means λ = 1/θ and P(T ≤ t) = 1 − e−t/θ:
Step 3Link 2 — "at least 2 of 10" given each maker
Given the maker, N ~ Bin(10, pi). Use the complement: at least 2 = not (0 or 1).
A shortcut that saves keystrokes: q1 = e−0.125, so q110 = e−1.25 = 0.286505 and q19 = e−1.125 = 0.324652.
Unpack this step: why q¹⁰ = e^(−1.25)
q = e−0.125, and a power of a power multiplies the exponents: (e−0.125)10 = e−1.25. One calculator press instead of ten multiplications, and no rounding build-up. (Meaning: all ten survive 1 500 h.)
Step 4Link 3 — Bayes
Enigma-2 was a 20% prior; the evidence (failures, which Enigma-2's weaker servers make more likely) raised it to about 37%.
1) Taking the mean as the rate: λ = 12 000 makes e−18 000 000 — nonsense. Rate = 1/mean. 2) Using "exactly 2 failed" instead of "at least 2". 3) Treating the 10 servers as ten independent draws of maker — "all belonging to the same manufacturer" means one maker per rack, so the maker is chosen once, at the root of the tree. 4) Reporting P(A | E2) = 0.7760 — that is the likelihood, not the posterior.
In-room check. P(E1 | A) = 0.265615/0.420820 = 0.6312, and 0.6312 + 0.3688 = 1 ✓. The posterior moved towards the maker whose likelihood was larger ✓.
Q5 · Cosmic particles on five detector squares · 3 + 5 marks §3.6, 3.4
Where this lives M2 notes · Poisson process (μ = rate × size) and binomial · M1 §2.5 · exactly k of n independent trials
Which tool(a) Poisson with μ = 2 × 1 = 2. (b) Each square is a yes/no trial ("empty?") with p = P(0 particles) — five independent trials, count the yeses: binomial built on top of a Poisson.
Step 1(a) Poisson, at most one
Let X = particles in one square, X ~ Poisson(2).
Step 2(b) Binomial on "empty"
p0 = P(X = 0) = e−2 = 0.135335. Let Y = number of empty squares; Y ~ Bin(5, 0.135335).
1) For (b), using the Poisson count over the whole 5 m² detector (μ = 10) — that answers "how many particles in total", not "how many squares are empty". 2) Forgetting C(5, 3): p03(1 − p0)2 is one particular arrangement of empty squares.
In-room check. Each square is empty only 13.5% of the time, so three empty out of five should be rare — 1.85% ✓.
Q6 · Two machines: Bayes three ways · 5 + 10 + 7 marks §2.4, 3.5, 3.4
🎧 Part (a) has a narrated walkthrough (about 5 min). The posterior is drawn as a share of an area, with three pause-and-commit questions.
Where this lives M1 §2.4 · total probability, Bayes · M2 · hypergeometric vs binomial
Which tool(a) Two-branch Bayes. (b) The same Bayes, but the evidence is "2 defective in 5 drawn without replacement from 100" — each branch's likelihood is a hypergeometric probability. (c) Binomial with p = the total probability from (a).
Step 1(a) Total probability, then Bayes
D = "chip defective". P(D) = 0.6 × 0.05 + 0.4 × 0.10 = 0.03 + 0.04 = 0.07.
Step 2(b) The shipment's contents, then hypergeometric likelihoods
A shipment of 100 from M1 contains 5 defective chips (5% of 100); from M2, 10. Let H = "exactly 2 defective among the 5 drawn". Drawing 5 of 100 without replacement:
"A machine is selected randomly": prior ½ each. The ½ and the C(100, 5) cancel in Bayes:
Unpack this step: C(95, 3) and C(90, 3) by hand
C(95, 3) = 95·94·93/6 = 138 415; C(90, 3) = 90·89·88/6 = 117 480. Cancel the 6 early: 93/3 = 31 and 94/2 = 47, so 95 × 47 × 31 = 138 415.
Step 3(c) Binomial with p = 0.07
Let X = defectives among 10; X ~ Bin(10, 0.07), the 0.07 from (a).
1) Binomial likelihoods in (b). "Without replacement" from a shipment of 100 is hypergeometric; n/N = 5/100 = 0.05 is right at the edge of the binomial rule of thumb, and the binomial shortcut gives 0.2272 instead of 0.2075 — a different answer at 4 dp. 2) Using 0.6 / 0.4 as the prior in (b). Those are production shares, the right prior for a random chip (part a). In (b) a machine is selected randomly: ½ each. With 0.6/0.4 you would get 0.2820. If you honestly read it the other way, write the assumption in one line — "assuming the machine is chosen with probability ½ each" — the paper asks you to define events, and a stated assumption earns method marks either way. 3) In (c), answering P(X = 2) or P(X < 2).
In-room check. (a) and (b) both say "defects point to M2": posterior of M2 goes 0.4 → 0.57 after one defective, and ½ → 0.79 after 2 of 5 ✓. (c) The mean is 0.7 defectives, so "at most 2" should be close to 1 ✓.
Paper 2 · 03 Mar 2025 · open book · 60 marks
“Answer all the questions. Symbols have their usual interpretation as per textbook. Marks for each question are mentioned at its end. Answer all parts of the same question together, showing all steps in detail. Define the event(s) and random variable(s) clearly. Simplify the answers fully. Specify all distributions properly.”
Open book that year, so no Φ values were printed; the values below are from Devore's Table A.3. In your closed-book paper, expect them printed.
Q1 · Exam times from an MGF · 9 marks §4.3 + R1 MGF
Where this lives M3 part 2 · MGFs of continuous distributions (normal MGF) · M3 notes §4.3 · standardising
Which toolRecognise the MGF → it is a normal; the one fact given (30.85% above 85) pins down σ; then two ordinary normal-table lookups.
Step 1Recognise the distribution
Multiply out the exponent: 0.5(σ²t² + 160t) = 80t + σ²t²/2. The normal MGF is eμt + σ²t²/2, and an MGF determines its distribution — so X ~ N(80, σ²): mean 80 minutes, σ still unknown.
Step 2Use the 30.85% to find σ
P(X > 85) = 0.3085 ⇒ P(Z > 5/σ) = 0.3085 ⇒ Φ(5/σ) = 0.6915. The table gives Φ(0.5) = 0.6915, so 5/σ = 0.5 and σ = 10.
Step 3(a) "Within one hour" = X ≤ 60
Step 4(b) Between 65 and 75
1) Reading σ² = 160 or μ = 160 from the exponent. Multiply the 0.5 in first: the t-coefficient is 80, and that coefficient is μ. 2) Solving Φ(5/σ) = 0.3085 — that is the area to the left; "more than 85" is the right tail. A negative σ is the alarm.
In-room check. 60 is 2σ below the mean: about 2.3% ✓. The band 65–75 is 1.5σ to 0.5σ below: about a quarter ✓.
Q2 · 900 students, fewer than 700 in favour · 4 marks §4.3
Where this lives M3 notes §4.3 · normal approximation, continuity correction
Which toolBinomial count, large n, "approximate" → normal with continuity correction. The same machine as 2026 Q3, other tail.
Step 1Define and check
X = number in favour, X ~ Bin(900, 0.8). np = 720, nq = 180, both ≥ 10 ✓. σ = √(900 × 0.8 × 0.2) = √144 = 12.
Step 2"Fewer than 700" = X ≤ 699 → 699.5
1) "Less than 700" corrected to 700.5 (that is "at most 700") gives 0.0521; no correction at 700 gives 0.0478. Write the integer event first: X ≤ 699. 2) σ = 144: that's the variance.
In-room check. 700 is 20 below a mean of 720, under 2σ: a few percent ✓. The exact binomial value is 0.0452, close to the approximation ✓.
Q3 · Battery life: gamma with α = 1 · 6 marks §4.4
Where this lives M3 notes §4.4 · exponential, memoryless
Which toolGamma with α = 1 is the exponential, with mean αβ = β. (b) is the memoryless property.
Step 1(a) Mean 20
Mean αβ = β = 20, so λ = 1/20: P(X ≤ 15) = 1 − e−15/20 = 1 − e−0.75 = 0.5276.
Step 2(b) Mean 7, memoryless
The 5 hours already used cancel out: the battery behaves as if new. Show the conditional-probability line, not just e−3/7 — the marks are for showing why.
1) Treating 20 as the rate: e−300 is the alarm. 2) In (b), computing P(Y ≥ 8) = 0.3189 — that ignores the information "still working after 5 hours". 3) Using the talk-time mean in (b); (b) is the internet-use battery.
Q4 · 30 units, 4 assigned: hypergeometric · 4 marks §3.5
Where this lives M2 notes · hypergeometric
Which toolSample without replacement from a pool with two kinds, count one kind: hypergeometric with N = 30, M = 20, n = 4.
Step 1Define and count the sample space
X = number of the 4 units needing a speed test, P(X = x) = C(20, x) C(10, 4 − x) / C(30, 4), and C(30, 4) = 30·29·28·27/24 = 27 405.
Step 2The three parts
1) Binomial with p = 2/3: 4/30 is far above 0.05, so the draws are not close to independent. 2) "Only 2" means exactly 2, not "at most 2".
In-room check. The missing value P(X = 1) = 20 × 120/27 405 = 2 400/27 405; the five numerators 210 + 2 400 + 8 550 + 11 400 + 4 845 = 27 405 ✓.
Q5 · Two coin types in an urn · 5 + 8 marks §2.4, 3.4
Where this lives M1 §2.4 · Bayes · M1 §2.5 · exactly k of n trials · Tutorial 2 Q5 from zero (Bayes after repeated tests)
Which toolBayes with priors 2/3 and 1/3. In (b) each likelihood is a binomial probability: exactly one head in three flips.
Step 1(a) One flip
Step 2(b) Likelihoods of "exactly one head in three"
Given the coin type, the three flips are independent, so the number of heads is Bin(3, p):
Step 3(b) Bayes
(Multiply top and bottom by 3 × 64 to clear both denominators.)
1) Dropping the 3 in 3pq². Here it cancels (it appears in both likelihoods), so the answer survives — but in a question with unequal flip counts it would not. Write it. 2) Priors ½ and ½ — there are two A coins and one B coin.
Q6 · A pmf recurrence · 12 marks §3.4–3.5
Where this lives M2 notes · the pmf ratio (new section) · M2 · the pmfs
Which toolDivide each pmf by itself one step earlier; the factorials cancel; rearrange the ratio into "α/(x+1) + β". For (b), multiply the recurrence by (x+1) and sum.
Step 1(a) Binomial
f(x) = C(n, x)pxqn−x. The ratio of the combinations is C(n, x+1)/C(n, x) = (n − x)/(x + 1), and the powers leave one extra p/q:
So α = (n+1)p/q, β = −p/q, with 𝒯 = {0, 1, …, n}. (At x = n the bracket is p/q − p/q = 0, correctly giving f(n+1) = 0.)
Unpack this step: why C(n, x+1)/C(n, x) = (n − x)/(x + 1)
C(n, x+1) = n!/[(x+1)! (n−x−1)!] and C(n, x) = n!/[x! (n−x)!]. Dividing: the n! cancels, x!/(x+1)! = 1/(x+1), and (n−x)!/(n−x−1)! = n − x.
Step 2(a) Negative binomial — Devore's convention
Devore's nb(x; r, p) = C(x+r−1, r−1)prqx, x = failures before the r-th success = 0, 1, 2, … . The same cancelling gives
So α = (r−1)q, β = q. Why it must be this convention: the question says X is non-negative integer-valued with the recurrence from 0 upwards; the "trials until the r-th success" version starts at r and has ratio xq/(x − r + 1), which is not of the given form. Say which convention you used in one line.
Step 3(b) The mean from the recurrence
Multiply both sides by (x+1): (x+1) f(x+1) = α f(x) + β(x+1) f(x). Sum over x = 0, 1, 2, …:
- Left: Σx≥0 (x+1) f(x+1) = Σy≥1 y f(y) = E(X) (put y = x+1; the y = 0 term would be 0 anyway).
- Right: α Σ f(x) + β Σ (x+1) f(x) = α · 1 + β(E(X) + 1).
Step 4(c) Negative binomial mean
α + β = (r−1)q + q = rq and 1 − β = 1 − q = p, so E(X) = rq/p = r(1 − p)/p — Devore's formula ✓.
1) Stopping at (n − x)p/((x+1)q) without splitting it into the asked form — the marks are for α and β. 2) Mixing conventions: using the trials-until pmf in (a) and then Devore's mean in (c).
In-room checks. The binomial through (b): α + β = np/q, 1 − β = (q + p)/q = 1/q, so E = np ✓ (the sum there is finite, but the same algebra works). Poisson: ratio μ/(x+1), i.e. α = μ, β = 0, giving E = μ ✓.
Q7 · Trials until the r-th success: MGF and a limit · 12 marks §3.5 + R1 MGF
Where this lives M2 notes · MGF · M3 part 2 · limits of MGFs · gamma and χ² in M3 notes §4.4
Which toolBuild the pmf by the "last trial is the r-th success" argument; sum the MGF series using the fact that Devore's negative binomial pmf sums to 1; then MaX(t) = MX(at) and a limit; match the result to a known MGF.
Step 1(a) The pmf
X = x means: trial x is a success, and exactly r − 1 of the first x − 1 trials are successes (in any positions).
Step 2(a) The MGF
Put k = x − r (the number of failures), so etx = etretk:
Devore's pmf sums to 1 for every 0 < q < 1: Σ C(k+r−1, r−1) qk = p−r = (1 − q)−r. The same identity with z = qet in place of q (valid while qet < 1, i.e. t < −ln q):
Check: r = 1 gives the geometric (trials) MGF in the Module 2 table ✓; M(0) = (p/p)r = 1 ✓.
Step 3(b) Scale: M of 2pX
Step 4(b) The limit as p → 0
Top and bottom both go to 0, so divide both by p first. The bottom is 1 − e2pt + pe2pt:
using e2pt → 1 and (1 − e2pt)/p → −2t. So
That is the gamma MGF (1 − βt)−α with α = r, β = 2 — which is chi-squared with ν = 2r degrees of freedom. Since MGFs converge, the distributions converge: for tiny p, 2pX is approximately χ²(2r).
Unpack this step: why (1 − e^(2pt))/p → −2t
The school limit (eh − 1)/h → 1 as h → 0 (it is the slope of ex at 0). With h = 2pt: (e2pt − 1)/p = 2t · (eh − 1)/h → 2t; flip the sign. L'Hôpital in p gives the same.
1) Using Devore's failures pmf here: the question counts trials, so x starts at r and the MGF has pet, not p, on top. 2) Writing MY(t) = 2p MX(t): scaling a variable scales the argument of the MGF. 3) Substituting p = 0 directly and getting 0/0 — divide by p first. 4) Stopping at "(1 − 2t)−r": the question says identify — name it, with parameters.
In-room check. Means: E(X) = r/p, so E(Y) = 2p · r/p = 2r for every p, and χ²(2r) has mean 2r ✓.
Paper 3 · 13 Mar 2024 · open book · 60 marks
“Answer all the questions. Symbols have their usual interpretation as per textbook. Marks to each question are mentioned at its end. Answer all parts of the same question together, showing all steps in detail. Define the event(s) and random variable(s) clearly. Simplify the answers fully. Specify all distributions properly.”
Two photos, pages 1 and 2. They carry someone's handwritten answers; all are checked below — all right except one intermediate line in Q3(b).
Q1 · Two shooters · 5 + 5 marks §2.4–2.5
Where this lives M1 §2.4 · conditional probability · M1 §2.5 · independence
Which toolDefinition of conditional probability, with the four outcomes' probabilities from independence (multiply).
Step 1The four outcomes
Let A = "X hits", B = "Y hits". Independent, so: both hit 0.7 × 0.4 = 0.28; only X 0.7 × 0.6 = 0.42; only Y 0.3 × 0.4 = 0.12; neither 0.3 × 0.6 = 0.18. (Sum 1 ✓.)
Step 2(a) Condition on "exactly one"
Step 3(b) Condition on "target hit"
"Target hit" = at least one hits = 1 − 0.18 = 0.82. "Y hit" is inside "target hit", so the intersection is just P(B) = 0.4:
1) In (b), using "only Y" (0.12) on top — the question asks whether Y hit, which includes "both hit". 2) Adding 0.7 + 0.4 = 1.1 for "target hit" — a probability above 1 means the overlap was double-counted; use 1 − P(neither).
Q2 · The gambler's two coins · 2 + 4 + 4 marks §2.4
Where this lives M1 §2.4 · Bayes · Tutorial 2 Q5 from zero (updating after repeated evidence)
Which toolBayes with priors ½, ½; the evidence is the whole sequence of flips so far.
Step 1(a) One head
F = fair, D = two-headed. P(H | F) = ½, P(H | D) = 1.
Step 2(b) Two heads
Same coin both times, so given the coin the flips are independent: P(HH | F) = ¼, P(HH | D) = 1.
Equivalently, update the answer to (a): prior ⅓, likelihood ½ vs 1 → (⅓ · ½)/(⅓ · ½ + ⅔ · 1) = ⅙/⅚ = 1/5. Both routes must agree.
Step 3(c) A tail
P(HHT | D) = 0: a two-headed coin cannot show tails. So P(F | HHT) = ½·⅛ / (½·⅛ + 0) = 1. One impossible observation settles it.
1) Treating the second flip as a new random choice of coin. "The same coin" means the coin is chosen once. 2) Reporting ½ for (c) because "the fair coin shows tails half the time" — that is a likelihood, not the posterior.
Q3 · A series: first team to i wins · 4 + (4 + 4) marks §3.4–3.5, 3.3
Where this lives M1 §2.5 · exactly k of n independent trials · M2 · negative binomial idea · M2 §3.3 · expected value
Which tool(a) Translate "seven games are played" into a statement about the first six. (b) List the possible values of the number of games (only 2 or 3), build its pmf, take E, then maximise in p with calculus.
Step 1(a) Seven games ⇔ 3–3 after six
A 7th game happens exactly when nobody has 4 wins after 6 games — the only way is 3 wins each. Whoever wins game 7 ends it, so game 7's result doesn't matter:
The route on the photo — "A wins in 7" plus "B wins in 7" — gives C(6, 3)[p4q3 + q4p3] = 20p3q3(p + q), which is the same thing since p + q = 1. The paper says "simplify fully" — finish at 20p3q3.
Step 2(b) The pmf of the number of games when i = 2
Let N = number of games. Two games suffice if the same team wins both (AA or BB); otherwise it is 1–1 and a third game decides. So
(Check: p² + q² + 2pq = (p + q)² = 1 ✓.)
Step 3(b) Expectation and its maximum
dE/dp = 2(1 − 2p) = 0 at p = ½; the second derivative is −4 < 0, a maximum. E = 2 + 2 · ¼ = 2.5 games. Evenly matched teams make the longest series — sensible.
The final line on the photo, 2(−p² + p + 1), and p = 0.5 are both right. But the line before it, "E(X) = 2/p + 2/(1 − p)", is not — at p = ½ it gives 8 games in a series that can last at most 3, and it has a minimum (not a maximum) at ½. It looks like the negative-binomial mean r/p borrowed for each team; that formula counts trials until the r-th success with no opponent stopping the series. Don't learn from that line. The in-room check that catches it: an expected count must lie between the smallest and largest values the count can take.
Q4 · Trains at a metro station · 2 + 2 + 2 + 2 marks §3.6, 4.4
Where this lives M2 · Poisson process (worked example 3: scale the rate first) · M3 §4.4 · exponential waiting times
Which toolPoisson process: the count in a window of length t is Poisson with mean μ = αt (rate × window). Get the rate into the question's unit first.
Step 1(a) Rate per minute, then the pmf
10 trains per 60 minutes is α = 10/60 = 1/6 train per minute. Let Xt = number of trains in t minutes. Then Xt ~ Poisson(t/6):
"Specify all distributions properly" — the range x = 0, 1, 2, … is part of the answer.
Step 2(b) and (c): put in the window
(b) t = 3, so μ = 3/6 = 0.5:
(c) t = 10, so μ = 10/6 = 5/3. "No trains" is x = 0, and μ0/0! = 1:
Step 3(d) "95% sure of at least one" — solve for t
At least one = everything except none: P(Xt ≥ 1) = 1 − e−t/6. We need this to be at least 0.95:
Take ln of both sides (ln is increasing, so the inequality keeps its direction): −t/6 ≤ ln 0.05 = −ln 20. Multiplying by −6 flips it: t ≥ 6 ln 20 = 6 × 2.99573 = 17.9744 minutes.
Same answer from Module 3: the wait for the first train is exponential with λ = 1/6, and P(wait ≤ t) = 1 − e−t/6 — it is the same event, "at least one train by time t".
1) Using μ = 10 for every window. The 10 is per 60 minutes; a 3-minute window has μ = 0.5. 2) For "at least one", summing P(1) + P(2) + … — it never ends; use 1 − P(0). 3) Losing the inequality in (d): multiplying by −6 flips ≤ to ≥. A time "at most 17.97 minutes" would mean shorter waits make you surer, which is backwards.
In-room check. Put t = 17.9744 back: e−2.9957 = 0.0500 ✓. And the mean gap between trains is 6 minutes, so needing about 3 mean gaps to be 95% sure is plausible.
Q5 · A tank of petrol · 4 + 4 marks §4.3
Where this lives M3 §4.3 · the normal distribution
Which toolStandardise, Z = (X − 60)/10, then one table lookup each. The only real work is translating the words into an inequality on X.
Step 1Name the variable, translate the words
Let X = distance (km) covered on one full tank, X ~ N(60, 10²).
(a) "Can cover at least 65 km" means the tank lasts 65 km or more: X ≥ 65.
(b) "Empty before covering 50 km" means the tank runs out short of 50: X < 50.
Step 2Standardise and look up
(For a continuous variable, < and ≤ give the same probability — no continuity correction here; that is only for a discrete count approximated by a normal.)
1) Reading (b) as X > 50 ("covers 50 km") — "empty before 50 km" is the car failing to reach 50. Write the inequality in words first. 2) Writing Φ(0.5) = 0.6915 as the answer to (a) — that is P(X ≤ 65); "at least" needs 1 − Φ.
In-room check. 65 is above the mean, so (a) must be below 0.5 ✓; 50 is further from the mean than 65, so (b)'s tail must be thinner than (a)'s ✓. (These are exactly the Φ(0.5) and Φ(1) values that were printed on the 2026 paper.)
Q6 · A two-piece pdf; an exponential solved for t · (3 + 3 + 3) + 3 marks §4.1–4.2, 4.4
Where this lives M3 §4.1–4.2 · pdf, cdf, expected value · M3 §4.4 · exponential
Which tool(a) Total area = 1, integrated piece by piece; then the cdf as a running area, one formula per range; then E(X) = ∫ x f(x) dx, again piece by piece. (b) Exponential cdf, then match exponents.
Step 1(a) Find c: the two areas must add to 1
So ⅛ + c + ⅛ = 1, giving c = ¾. Legitimate also needs f ≥ 0: on [0, 1) it runs from ¾ to 1, on [1, 2] from ¼ down to 1/16 — never negative ✓.
Step 2(a) The cdf, one range at a time
F(x) = ∫−∞x f = area to the left of x. On [0, 1): ∫0x (u/4 + ¾) du = x²/8 + 3x/4 = (x² + 6x)/8. At x = 1 that is 7/8 — all the area of the first piece. On [1, 2] add the second piece's area so far: 7/8 + ∫1x 1/(4u²) du = 7/8 + ¼ − 1/(4x) = 9/8 − 1/(4x).
Checks: both formulas give 7/8 at x = 1 (a cdf of a continuous variable has no jumps) ✓; F(2) = 9/8 − 1/8 = 1 ✓.
Step 3(a) E(X), piece by piece
The second piece: x · 1/(4x²) = 1/(4x), and ∫ dx/x = ln x, so ¼(ln 2 − ln 1) = ¼ ln 2. Numerically 0.45833 + 0.17329 = 0.6316.
Step 4(b) Exponential, solved for t
Let T = rainfall duration (hours). Mean 2 means λ = 1/2, so P(T < t) = 1 − e−t/2. Setting this equal to 1 − e−1 gives e−t/2 = e−1, so t/2 = 1, t = 2 hours. The idea behind it: for any exponential, P(T < mean) = 1 − e−1 ≈ 0.6321.
1) Stating the cdf only on [0, 1) and [1, 2]. A cdf is defined for every real x — write the "0 for x < 0" and "1 for x > 2" lines; they carry marks. 2) Forgetting the 7/8 carried in on [1, 2] and writing ¼ − 1/(4x) — that curve starts at 0 at x = 1, and the jump shows it's wrong. 3) Integrating 1/(4x) as a power (x0/0) — it is the one power that gives a log. 4) In (b), taking λ = 2 from "average 2": the mean is 1/λ.
In-room check. E(X) = 0.6316 lies inside [0, 2] and below 1, which fits: 7/8 of the probability sits on [0, 1).
Every answer on this page of the photo is right: 0.075816, 0.1888756, 17.97439 min, 0.3085, 0.1587, c = 0.75, the cdf pieces ((x+3)² − 9)/8 and 1.125 − 0.25/x, E = 0.63162, t = 2. One thing to add in the exam: the cdf answer should list all four ranges (including 0 below and 1 above), and round to the four decimal places the paper asks for.
All answers, one table
| Paper · Q | Family | Answer (4 dp) |
|---|---|---|
| 2026 Q1 | E from a cdf | 1/3 = 0.3333 |
| 2026 Q2 | E of a function, then minimise | min 0.2500 at g = 0.5 |
| 2026 Q3 | Normal approx., continuity correction | 0.0606 |
| 2026 Q4 | Exponential → binomial → Bayes | 0.3688 |
| 2026 Q5 | Poisson; binomial on "empty" | 0.4060 · 0.0185 |
| 2026 Q6 | Bayes · hypergeometric Bayes · binomial | 0.5714 · 0.2075 · 0.9717 |
| 2025 Q1 | Recognise the normal MGF, solve for σ | σ = 10 · 0.0228 · 0.2417 |
| 2025 Q2 | Normal approx., continuity correction | 0.0436 |
| 2025 Q3 | Exponential, memoryless | 0.5276 · 0.6514 |
| 2025 Q4 | Hypergeometric | 0.0077 · 0.3120 · 0.5928 |
| 2025 Q5 | Bayes; binomial likelihood | 0.4000 · 0.1429 |
| 2025 Q6 | pmf recurrence | see Q6; nb mean r(1−p)/p |
| 2025 Q7 | nb (trials) MGF; MGF limit | [pet/(1−qet)]r · χ²(2r) |
| 2024 Q1 | Conditional, independent events | 0.2222 · 0.4878 |
| 2024 Q2 | Sequential Bayes | 0.3333 · 0.2000 · 1 |
| 2024 Q3 | Series counting; E and maximise | 20p³q³ · 2(1+p−p²), max 2.5 at p = 0.5 |
| 2024 Q4 | Poisson process; solve for t | 0.0758 · 0.1889 · 17.9744 min |
| 2024 Q5 | Normal tails | 0.3085 · 0.1587 |
| 2024 Q6 | Piecewise pdf: c, cdf, E · exponential solve for t | c = 0.75 · E = 0.6316 · t = 2 |
Every number above was recomputed independently (exact binomial sums checked against the normal approximations; the series answers checked by enumerating all game sequences). Now drill the same families with fresh numbers: mid-sem drill.
"Devore §4.4 and §2.4: a rack of 10 servers comes from maker 1 (prior 0.8, exponential lifetimes with mean 12 000 h) or maker 2 (prior 0.2, mean 5 000 h); after 1 500 h at least 2 have failed. Walk me through the three links — failure probability, binomial likelihood, Bayes — stopping after each so I can do the next number myself."
"Devore §4.3: give me five normal-approximation-to-binomial questions like 'P(at least 525 heads in 1000 fair tosses)', mixing 'fewer than', 'at most', 'at least', 'more than'. For each, ask me first only for the continuity-corrected boundary, and check it before I standardise."
One caution: AI answers can contain confident arithmetic slips — recompute every final number, and check table values against Devore's Table A.3.